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# GCSE - Arithmetic Sequence watch

1. Hi, I'd appreciate some help with this question. I'm not sure where to start.

The first five terms of an arithmetic sequence are:
x+1, 2x, 2(2x+3)/6-x, x^2-2, 5x-3
Show that the term 4x^2-3 is not in the sequence.
2. it is barely conceivable that this is a GCSE question

smh

anyhow....

you can find a rule in the usual way;

the gap between the first two terms is x - 1

so (x - 1) *N + ?

letting N = 1 we can see that ? must be 2

so the nth term is (x - 1)*N + 2

see what happens if you let 4x2 - 3 equal (x - 1)*N + 2
3. (Original post by the bear)
it is barely conceivable that this is a GCSE question

smh

anyhow....

you can find a rule in the usual way;

the gap between the first two terms is x - 1

so (x - 1) *N + ?

letting N = 1 we can see that ? must be 2

so the nth term is (x - 1)*N + 2

see what happens if you let 4x2 - 3 equal (x - 1)*N + 2
Okay, thanks. Is this right?

Just using a term in the sequence to show whether I know a term fits:
2x=(x-1)n+2
2x-2=(x-1)n
2(x-1)=(x-1)n
So n=2, so fits the sequence.

In the case of 4x^2-3:
4x^2-3=(x-1)n+2
4x^2-5=(x-1)n
4x^2-5 cannot be factorised to give an (x-1) bracket, so is not divisible by it and does therefore not fit the sequence.
4. (Original post by irrelevant kid)
Okay, thanks. Is this right?

Just using a term in the sequence to show whether I know a term fits:
2x=(x-1)n+2
2x-2=(x-1)n
2(x-1)=(x-1)n
So n=2, so fits the sequence.

In the case of 4x^2-3:
4x^2-3=(x-1)n+2
4x^2-5=(x-1)n
4x^2-5 cannot be factorised to give an (x-1) bracket, so is not divisible by it and does therefore not fit the sequence.
you would need to show a bit more detail of "4x^2-5 cannot be factorised to give an (x-1) bracket", but your idea is very good.
5. (Original post by the bear)
you would need to show a bit more detail of "4x^2-5 cannot be factorised to give an (x-1) bracket", but your idea is very good.
Okay, thanks a lot!
6. (Original post by irrelevant kid)
Okay, thanks a lot!
are you sure it was a GCSE question ?
7. (Original post by the bear)
are you sure it was a GCSE question ?
Yeah, my teacher gave me a booklet of grade 9+ questions today and this was one of them. Most of the questions weren't too bad, but this one took me a while.
8. just realized that you can find x quite easily... there is only one suitable value

9. (Original post by the bear)
just realized that you can find x quite easily... there is only one suitable value

The question is wanting to work out the value of x first. Then once obtained the value of x it wants you to substitute that value in to form a sequence which you must work out the nth term for. Now you must work out the value of 4x²−3 and show that it isn’t in the pattern of the nth term
10. (Original post by irrelevant kid)
Okay, thanks. Is this right?

Just using a term in the sequence to show whether I know a term fits:
2x=(x-1)n+2
2x-2=(x-1)n
2(x-1)=(x-1)n
So n=2, so fits the sequence.

In the case of 4x^2-3:
4x^2-3=(x-1)n+2
4x^2-5=(x-1)n
4x^2-5 cannot be factorised to give an (x-1) bracket, so is not divisible by it and does therefore not fit the sequence.
This is incorrect
11. (Original post by Y11_Maths)
This is incorrect
it is correct, but not the best method.
12. (Original post by the bear)
it is correct, but not the best method.
You won’t obtain any marks for this method though because it is not what the question is asking for
13. (Original post by Y11_Maths)
You won’t obtain any marks for this method though because it is not what the question is asking for
the question does not specifically ask you to find x.
14. (Original post by the bear)
the question does not specifically ask you to find x.
It is a 6 mark question so I would edge my bets. I posted how the question is meant to be answered above and I have answered this question before and with all the steps it takes it definitely wants you to find x and solve.
15. (Original post by Y11_Maths)
It is a 6 mark question so I would edge my bets. I posted how the question is meant to be answered above and I have answered this question before and with all the steps it takes it definitely wants you to find x and solve.
if you still think the other method does not work please explain why.
16. (Original post by the bear)
if you still think the other method does not work please explain why.
I didn’t say it doesn’t work I’m saying it is incorrect as you would not obtain any marks from it since that is not how the question is meant to be answered.
17. (Original post by irrelevant kid)
Okay, thanks. Is this right?

Just using a term in the sequence to show whether I know a term fits:
2x=(x-1)n+2
2x-2=(x-1)n
2(x-1)=(x-1)n
So n=2, so fits the sequence.

In the case of 4x^2-3:
4x^2-3=(x-1)n+2
4x^2-5=(x-1)n
4x^2-5 cannot be factorised to give an (x-1) bracket, so is not divisible by it and does therefore not fit the sequence.
Just because 4x^2-5 can't be factored to give an x-1 bracket, it doesn't mean that x-1 doesn't divide 4x^2-5.

For example consider x-1 and x+1. Clearly x+1 doesn't contain an x-1 bracket. But this doesn't necessarily mean x-1 isn't a factor.

For example, if x=3 then: x-1=2, x+1=4. Obviously 2 is a factor of 4.
18. (Original post by psc---maths)
Just because 4x^2-5 can't be factored to give an x-1 bracket, it doesn't mean that x-1 doesn't divide 4x^2-5.

For example consider x-1 and x+1. Clearly x+1 doesn't contain an x-1 bracket. But this doesn't necessarily mean x-1 isn't a factor.

For example, if x=3 then: x-1=2, x+1=4. Obviously 2 is a factor of 4.
Thank you
19. (Original post by Y11_Maths)
This is incorrect
Is this right?

2x - (x+1)
Difference = x-1
5x-3-(x^2-2)
Difference = 5x-x^2-1

x-1=5x-x^2-1
x^2-4x=0
x(x-4)=0
x=4

x+1=5
2x=8
2(2x+3)/6-x=11
x^2-2=14
5x-3=17
nth term= 3n+2

4x^2-3 = 61
3n+2=61
3n=59
59/3 is not an integer so 4x^2-3 is not a term in the sequence.
20. (Original post by irrelevant kid)
Is this right?

2x - (x+1)
Difference = x-1
5x-3-(x^2-2)
Difference = 5x-x^2-1

x-1=5x-x^2-1
x^2-4x=0
x(x-4)=0
x=4

x+1=5
2x=8
2(2x+3)/6-x=11
x^2-2=14
5x-3=17
nth term= 3n+2

4x^2-3 = 61
3n+2=61
3n=59
59/3 is not an integer so 4x^2-3 is not a term in the sequence.
I different method to mine but yes this is correct well done

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